Given functions $A(x), B(x)$ find $f(x)$ s.t. $A\big(f(x)\big)=f\big(B(x)\big)$

Currently, I am facing this problem:

Given two real functions $$A( \vec x )$$ and $$B( \vec x ):\Bbb R^N\to \Bbb R$$, I want to find a third real, monotonic function $$f(x):\Bbb R\to\Bbb R$$ such that:

$$A\big(f( \vec x )\big)=f\big(B( \vec x )\big)$$

where the simplified the notation writing $$f( \vec x )$$ means $$f(x_1,x_2,\ldots,x_n) = \big(f(x_1), f(x_2),\ldots, f(x_n)\big).$$

I am interested in either having a formula/method for finding $$f$$, or even just having a proof that $$f$$ exists (or doesn't) under some specific conditions. Eventually, I am interested also in the solution in the case $$N=1$$.

Also, does this type of problem have a specific name?

Thank you very much!

• Fixed! I hope now it's fun enough – SchroedingerDidStuff Feb 10 at 9:37
• you might also want to consider changing the name of your profile; the adjectives you use to characterise Schrödinger are inappropriate, IMO. – Carlo Beenakker Feb 10 at 9:49
• Ok, I also changed the nick to a more appropriate one. I guess the next answer I'll receive is not gonna be on the question I asked, but on the way, I use the comma. Maybe I should ask this same question on a grammar forum... – SchroedingerDidStuff Feb 10 at 9:58
• It seems too easy: if $A(x_1,x_2)=1+x_1$ and $B(x_1,x_2)=x_1+x_2$ then you are looking for $f$ such that $1+f(x_1)=f(x_1+x_2)$ which cannot be, for example if $x_2=0$. Maybe you missed some conditions? (Or did I?) – Yaakov Baruch Feb 10 at 10:09
• Also, does this type of problem have a specific name? Given two maps $A$ and $B$, trying to find an $f$ such that $A \circ f =f \circ B$ is a problem of semi-conjugacy. – Laurent Berger Feb 10 at 13:12

Existence of such $$f$$ is a very strong condition on $$A$$ and $$B$$ which is called semi-conjugacy. Of course, for generic $$A$$ and $$B$$ function $$f$$ does not exist. Suppose for simplicity that $$N=1$$. Then it is clear that the image of any fixed point of $$B$$ under $$f$$ is a fixed point of $$A$$. Furthermore, when $$N=1$$, your equation implies that $$A^n\circ f=f\circ B^n$$ where $$A^n$$ means the $$n$$-th iterate. This implies that periodic points of $$B$$ are mapped to periodic points of $$A$$ of the same period. So we have some very complicated relation between $$A$$ and $$B$$.